MAP 2302 Final Exam Review

Instructions: The final exam will consist of 6 questions, although some questions may consist of multiple parts, and be worth 150 points. Be sure to show as much work as possible in order to demonstrate that you know what you are doing. The point value for each question is listed after each question. A scientific calculator may be used but no graphing calculators or calculators on any device (cell phone, iPod, etc.) which can be used for any other purpose. The concepts and types of questions on the final exam will be similar to the previous tests, previous reviews, and this review, although the numbers and functions may be different on each question on the exam. If you wish to use the list of Laplace Transforms or Common Taylor Series I have provided online you may do so, but you must bring your own copy as I will not provide it on the test and sharing is not allowed. Be sure that there is no writing on the list. Each of the 6 questions can be found in the following places:

1. Exam #1, Problem #4 (15 points)

2. Exam #1, Problem #5 (15 points)

3. Exam #2, Problem #3 (20 points)

4. Exam #3, Problem #4 (40 points)

5. Exam #4, Problem #3 (25 points)

6. Exam #4, Problem #4 (25 points)

(1) Find a general explicit solution of the differential equation. (15 points)

~dy _ y

xdx JX2 - 16

af -' Ju - J x "CAAJ J"J - m

U:. x2.- fl"

etu-=- ZXdJx => ~ d\A -= XM

(2) Find the general solution to the differential equation. Primes denote derivatives with respect to x. (15 points)

y' + (tan x) y = sec x

f (x) ~ +un X

SrrkJ AA ::: r-\"" ~ clJ)( -:0 f u~U>~~

ettA~ - &lY\"(. olh. ~ -ellA ~ '011\ ~ok

-j i Ju. :: -I" I.A ::. - I", COS l<:

\~ (tO~ leY' I", ~( X - e - .e - ~( X

£tl X~ I t-{~ )c') y) -=- ~( X·<;lCX

(~t 'Ie •J')' -=- ~L'2.'K

fC~t '"·lJx )K~JJx00 ~ Stc XOj :: ~ X +L,

J':.. COS , ~x~ C·c.oSXo

4-:..~ o ~ t C·LobX J c~

(3) Find the general solution of each differential equation. (20 points)

y" - 5y' + 4y = 8ex + cos x

1II-5yI

~/~O

W12. -~(Y\ t-y :::: 0

"tx X ~c. =C,e +Cl!/

GIfi~ : yp '; ~e)( +B~ ~ -I- CSi" )(

()..~ G~s~ ~ ~p ~ A-x~ 'It +8cosX+ CS1AX

x~ r' ~ ~t)( ... )e ) - e, '0"" ):. ~ Cc.(6)c

~p' '= I'{e~ \ [u;')l1- ~ -C.~i"l<

:::. A(l\- 2\ ~x. ~~toSK- C~i(\ 'x

~r' -r;r~ f ~~f=Afj;;)l-~~ -~~i~

-'5(A~~)i-~-~os~l ~ ~[~{ t&»sx +~] ::: - :)Ae.X + (SB- 3C}S fA 'Ie ... ( 3B-I-5C )co5 X ::...%/ ·Hos X

-?>A ='<6 ~A = - 3 ~

s(5B-3C=O) ~ 2~-1 c.~o

~(-:oBtSL -== \) ~ q & +--1 L ::3 .l..

3'1 B-="3 ~ i3>:: 3\f

(4) Use Laplace transforms to solve the initial value problem. (20 points each)

(a) y' - y = 2cos5t; y(O) = 1

\ d-.S +-.L y(~) -=- C~-1\(~lH~) <;-\

A-(c:'l.~~S) t (Ib~+() (~-i)

(~ -i) (<;. ~ ~ L S')

~S-::: (\ (s~~S) t (8{;t C) (S-1) -L

S-=-1'. ~ ~ ;).CtJ A =:; A-:; \~

1~ ~ -k (~l.VLC:;) 4- ~~1 +-(C- &)~-G

Js -:: (~ -h\~1 ~ LC-e)~+ (W -C) \ ,

~"'"13 :: 0 ~&-:: - li

c-e, -==-a

2t(b) y" - 4y' + 4y = t3e ; y(O) = 0, y'(O) = 0

f~~"J - ~t\J'}+~ t~j~O t~t3t:lt}

tf Y(~) - st-~-'i (SYlI)-t)) tl{ ,II) 0= (s_~l", o V 0

s"l-YL~) -\{C; y(~) t-4 Y()l -=- CS~L)~

y(~\ CSZ-~~+'1) -=- (s -~) i(

Y(\)(s-i)" -:. C~~\~ 5 ·Lf · 3·1·\

( C; -'2.) S"t!

(5) Use the power series method to solve the given initial-value problem. Identify the series solution in terms of familiar elementary functions. (25 points)

y" + 4y = 0; y(O) = 0, y'(O) = 3

00

Y-=- r.. ~ -Xk: :: ~+C X: +C )(1 +C x"\ . - . t.~ I 2 "S

~'(())=C,:::3

'1" =- t k(l -\)C~Xl-2 ( k--:: l

!')o . ~. 2 /)I ~

L.. kU~. -I) C~X +- ~ 'LCLX :: 0 ~ ~=-o

f.. (~H)[t41)C~"1X~ l:-~

-- -

(5 ) (continued)

'-tCoC2 -=- =D

J · t

'{c. \\:-3Cl> -=

3 '2 ~

ttc z__C ~ oIt 4'~ ­

C -- ~ c:) - ~ . y . ~ s- ....-s 'Y 5 ·'-{· yZ­

~cC ­lD - (, . ~ • := 0

yesC-=t :: . ~ . Y·4 · CL.. _ ~.~ t · ro · S-·~ · l ·"l

(6) Find two power series solution of the given differential equation. Write out the first three terms of each series solution. (25 points)

(x + l)y" - (2 - x)y' + y = 0 00

1 ~ ~C~'(k. , 'I ' ='£ kC"-X l -,, yet -==f k(~-I)C X\c-2 a:.. ~ v I ~-=- . ~=z. It-

Po ~ ~2 00 l - l lt;JC

(~t(} L k(t'1)C~X - (L~X)L lCt-~ +- t. Ll ~ ::- 0 k.:~1. ~=-I J':..t-Q

00 ( \ l--I Dc _ Jlc 1)0 k.. Do k. ~ k ~-f)elX +L ll~-j)l~X 1_ 2L ~L~l~~1 +- L. ~~X +-~ Cit.'(- 0 ~~ ~~-=-t ~ &k.r~ ""Itt. Xl S-hub wit{,. ",0 s-lam 1Jl~)o;0 ~.xl S1t;,rts wi'tt.. ~o

~Cl -C,-I-W~ t l ~(~lCl"} (k.nl(L~\)C~.-.l[u~~ +-kc". +-~1l :c 0 ~-=-t

Je~ - (I +-Cv :: \)

'(k.~i)(lt\\C. +- Ll-2.)(~+1) C.. +(tt1) C~ <:-0Ln. ~+ I

(6) (continued)

-

Scratch Paper

Js +- -Ly()) :: G--tl(~'HS) 5-1

\ '2~ - -3;- (3

I I~ -L- ---- ..\- +- I~ 5- ' S~ +-2)" S-l